Explanation Given information: The relation R is an equivalence relation on the set A. A = { 1 , 2 , 3 , 4 , ... , 20 } . R is defined on A as follows: For all x , y ∈ A , x R y ⇔ 4 | ( x − y ) . Calculation: A = { 1 , 2 , 3 , 4 , … , 20 } R = { ( x , y ) ∈ A × A | 4 ( x − y ) } Let us first group the ordered pairs in R that have at least one common element (with at least one of the other elements in the group). 4 | ( x − y ) is true if and only if x − y is a multiple of 4 if and only if x and y differ by 0, 4, 8, 12 or 16 (as the elements are between 1 and 20). Thus ( x − y ) ∈ R when x and y differ by 0, 4, 8, 12 or 16. First group: ( 1 , 1 ) , ( 1 , 5 ) , ( 1 , 9 ) , ( 1 , 13 ) , ( 1 , 17 ) ( 5 , 1 ) , ( 5 , 5 ) , ( 5 , 9 ) , ( 5 , 13 ) , ( 5 , 17 ) ( 9 , 1 ) , ( 9 , 5 ) , ( 9 , 9 ) , ( 9 , 13 ) , ( 9 , 17 ) ( 13 , 1 ) , ( 13 , 5 ) , ( 13 , 9 ) , ( 13 , 13 ) , ( 13 , 17 ) ( 17 , 1 ) , ( 17 , 5 ) , ( 17 , 9 ) , ( 17 , 13 ) , ( 17 , 17 ) Second group: ( 2 , 2 ) , ( 2 , 6 ) , ( 2 , 10 ) , ( 2 , 14 ) , ( 2 , 18 ) ( 6 , 2 ) , ( 6 , 6 ) , ( 6 , 10 ) , ( 6 , 14 ) , ( 6 , 18 ) ( 10 , 2 ) , ( 10 , 6 ) , ( 10 , 10 ) , ( 10 , 14 ) , ( 10 , 18 ) ( 14 , 2 ) , ( 14 , 6 ) , ( 14 , 10 ) , ( 14 , 14 ) , ( 14 , 18 ) ( 18 , 2 ) , ( 18 , 6 ) (

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Chapter 8.3, Problem 5ES

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To find the distinct equivalence classes of R.

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Chapter 8 Solutions

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Ch. 8.1 - If R is a relation from A to B, xA , and yB , the...Ch. 8.1 - Prob. 2TY Ch. 8.1 - Prob. 3TY Ch. 8.1 - Prob. 4TY Ch. 8.1 - If R is a relation on a set A, the directed graph...Ch. 8.1 - As in Example 8.1.2, the congruence modulo 2...Ch. 8.1 - Prove that for all integers m and n,m-n is even...Ch. 8.1 - The congruence modulo 3 relation, T, is defined...Ch. 8.1 - Define a relation P on Z as follows: For every...Ch. 8.1 - Prob. 5ES

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To find the distinct equivalence classes of R.

Solution Summary: The author explains that R is an equivalence relation on the set A. They first group the ordered pairs in R that have at least one common element.

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